I am interested in testing whether being incarcerated changes opiod use in at-ri
ID: 3050522 • Letter: I
Question
I am interested in testing whether being incarcerated changes opiod use in at-risk adolescents. I know that on average, at-risk adolescents use opiods 6 times a month (sd 3.4). I survey a sample of 15 adolescents in juvenile detention facilities in the state of Colorado and find that on average, they use opiods 8 times a month.
1) Calculate the z-score (for a sampling distribution) to test if adolescents in juvenile detention use significantly more opiods than the general at-risk population. Show your work.
2)Using that z-value and the normal distribution applet (http://byuimath.com/apps/normprob.html), calculate the area to the left of the negative and right of the positive z-value on the applet. Add them up. What do you get? This number is the p-value of your z-score for a two-tailed test. Alternatively, you may use the R code provided to calculate your p-value. If you choose to do this, please provide your input code and output!
3)Given this z-score and p-value, would you reject or fail to reject the null hypothesis under a significance level of .05?
4)Interpret the meaning of your p-value.
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: < 6
Alternative hypothesis: > 6
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.
Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).
SE = s / sqrt(n)
S.E = 0.8779
z = (x - ) / SE
z = 2.28
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a z statistic test statistic of 2.28.
2) Thus the P-value in this analysis is = 0.0113
3) Interpret results. Since the P-value (0.0113) is less than the significance level (0.05), we have to reject the null hypothesis.
4) The p-value (0.0113) means that the chance that the juvenile detention use significantly more opiods than the general at-risk population is 0.0113.
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